# Law of cosines

(Difference between revisions)
 Revision as of 09:53, 3 June 2011 (edit)← Previous diff Revision as of 10:25, 3 June 2011 (edit) (undo)Next diff → Line 20: Line 20: ==Proof== ==Proof== ===By the Pythagorean Theorem=== ===By the Pythagorean Theorem=== - An easy way to think of the law of cosines is as an extension of the Pythagorean theorem: + An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle: ::$a^{2} + b^{2} = c^{2}$ ::$a^{2} + b^{2} = c^{2}$ - [[Image:Pythagorean_cosines_proof.jpg]] + [[Image:Pythagoras_law_of_cosines.jpg]] + By Pythagorean theorem, we know + ::$c^{2} = p^{2} + h^{2}$ + + But $p$ is just some portion of side length $a$ which is $r$ less than the length of $a$. Substituting the difference gives us, + + ::$c^{2} = (a-r)^{2} + h^{2}$ + + By Pythagorean theorem, we also know that + + ::$h^{2} = b^{2} - r^{2}$ + + Substituting the appropriate values gives us, + + ::$c^{2} = (a-r)^{2} + ( b^{2} - r^{2})$ + + Expanding the squared term gives us + + ::$c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}$ + + Simplify for + + ::$c^{2} = a^{2} -2ar + b^{2}$ + + And by the definition of cosine, we know that + + ::$r=b \cos C$ + + Substituting this value in give us + + ::$c^{2} = a^{2} -2ab \cos C + b^{2}$ + + or + + ::$c^{2} = a^{2}+ b^{2} -2ab \cos C$ ===Using the Distance Formula=== ===Using the Distance Formula=== Line 34: Line 68: - $distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ + ::$\text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ $c$ is the distance from $A$ to $B$. $c$ is the distance from $A$ to $B$. Line 40: Line 74: Substituting the appropriate points into the distance formula gives us Substituting the appropriate points into the distance formula gives us - $c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$ + ::$c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$ Squaring the inner terms, we have Squaring the inner terms, we have - $c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$ + ::$c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$ Since $\cos^{2} C + \sin^{2} C = 1$, Since $\cos^{2} C + \sin^{2} C = 1$, - $c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}$ + ::$c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}$ Square both sides for Square both sides for - $c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}$ + ::$c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}$ ==Example Problem== ==Example Problem== Line 58: Line 92: - $c^{2} = a^{2} + b^{2} - 2ab \cos C$ + ::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ [[Image:SAS triangle.jpg]] [[Image:SAS triangle.jpg]] Line 67: Line 101: - $c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$ + :$c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$ Simplify for Simplify for - $c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$ + ::$c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$ Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us - $c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$ + ::$c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$ Simplify for Simplify for - $c^{2} =36 + 72 - 72$ + ::$c^{2} =36 + 72 - 72$ - $c^{2} =36$ + ::$c^{2} =36$ Taking the square root of both sides gives us Taking the square root of both sides gives us - $c =6$ + ::$c =6$ ---- ---- Line 93: Line 127: - $b^{2} = a^{2} + c^{2} - 2ab \cos B$ + ::$b^{2} = a^{2} + c^{2} - 2ab \cos B$ Substituting in the appropriate side lengths gives us Substituting in the appropriate side lengths gives us - $(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ + ::$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ Simplify for Simplify for - $36 (2) = 36 + 36 - 72 \cos B$ + ::$36 (2) = 36 + 36 - 72 \cos B$ - $72 = 72 - 72 \cos B$ + ::$72 = 72 - 72 \cos B$ Subtracting $72$ from both sides gives us Subtracting $72$ from both sides gives us - $0 = - 72 \cos B$ + ::$0 = - 72 \cos B$ Dividing both sides by $-72$ gives us Dividing both sides by $-72$ gives us - $0 = \cos B$ + ::$0 = \cos B$ Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that Using [[Basic Trigonometric Functions#Inverse Trig Functions| inverse trigonometry]], we know that - $B = 90^\circ$ + ::$B = 90^\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$ - $180^\circ - 90^\circ - 45^\circ = 45^\circ$ + ::$180^\circ - 90^\circ - 45^\circ = 45^\circ$ - $A=45^\circ$ + ::$A=45^\circ$ [[Image:SAS_solution.jpg]] [[Image:SAS_solution.jpg]]

## Revision as of 10:25, 3 June 2011

Law of Cosines

The law of cosines is a trigonometric extension of the Pythagorean Theorem.

# Basic Description

The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$

When to use it: SAS, SSS.

# A More Mathematical Explanation

## Proof

### By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extens [...]

## Proof

### By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle:

$a^{2} + b^{2} = c^{2}$

By Pythagorean theorem, we know

$c^{2} = p^{2} + h^{2}$

But $p$ is just some portion of side length $a$ which is $r$ less than the length of $a$. Substituting the difference gives us,

$c^{2} = (a-r)^{2} + h^{2}$

By Pythagorean theorem, we also know that

$h^{2} = b^{2} - r^{2}$

Substituting the appropriate values gives us,

$c^{2} = (a-r)^{2} + ( b^{2} - r^{2})$

Expanding the squared term gives us

$c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}$

Simplify for

$c^{2} = a^{2} -2ar + b^{2}$

And by the definition of cosine, we know that

$r=b \cos C$

Substituting this value in give us

$c^{2} = a^{2} -2ab \cos C + b^{2}$

or

$c^{2} = a^{2}+ b^{2} -2ab \cos C$

### Using the Distance Formula

Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$.

$\text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

$c$ is the distance from $A$ to $B$.

Substituting the appropriate points into the distance formula gives us

$c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$

Squaring the inner terms, we have

$c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$

Since $\cos^{2} C + \sin^{2} C = 1$,

$c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}$

Square both sides for

$c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}$

## Example Problem

Complete the triangle using the law of cosines.

$c^{2} = a^{2} + b^{2} - 2ab \cos C$

### Solution

To find the side length $c$,

$c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$

Simplify for

$c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$

Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us

$c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$

Simplify for

$c^{2} =36 + 72 - 72$

$c^{2} =36$

Taking the square root of both sides gives us

$c =6$

Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$,

$b^{2} = a^{2} + c^{2} - 2ab \cos B$

Substituting in the appropriate side lengths gives us

$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$

Simplify for

$36 (2) = 36 + 36 - 72 \cos B$
$72 = 72 - 72 \cos B$

Subtracting $72$ from both sides gives us

$0 = - 72 \cos B$

Dividing both sides by $-72$ gives us

$0 = \cos B$

Using inverse trigonometry, we know that

$B = 90^\circ$

And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$

$180^\circ - 90^\circ - 45^\circ = 45^\circ$
$A=45^\circ$

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